Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $x = \dfrac{p^2 + p - 90}{-7p + 63} \times \dfrac{4p - 12}{p + 10} $
Explanation: First factor the quadratic. $x = \dfrac{(p + 10)(p - 9)}{-7p + 63} \times \dfrac{4p - 12}{p + 10} $ Then factor out any other terms. $x = \dfrac{(p + 10)(p - 9)}{-7(p - 9)} \times \dfrac{4(p - 3)}{p + 10} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ (p + 10)(p - 9) \times 4(p - 3) } { -7(p - 9) \times (p + 10) } $ $x = \dfrac{ 4(p + 10)(p - 9)(p - 3)}{ -7(p - 9)(p + 10)} $ Notice that $(p - 9)$ and $(p + 10)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ 4\cancel{(p + 10)}(p - 9)(p - 3)}{ -7(p - 9)\cancel{(p + 10)}} $ We are dividing by $p + 10$ , so $p + 10 \neq 0$ Therefore, $p \neq -10$ $x = \dfrac{ 4\cancel{(p + 10)}\cancel{(p - 9)}(p - 3)}{ -7\cancel{(p - 9)}\cancel{(p + 10)}} $ We are dividing by $p - 9$ , so $p - 9 \neq 0$ Therefore, $p \neq 9$ $x = \dfrac{4(p - 3)}{-7} $ $x = \dfrac{-4(p - 3)}{7} ; \space p \neq -10 ; \space p \neq 9 $